Factor using the ac-method.
-$15 y ^ { 4 } + 2 y ^ { 2 } - 8$
A) $\left( 15 y ^ { 2 } - 2 \right) \left( y ^ { 2 } + 4 \right)$
B) $\left( 3 y ^ { 2 } - 2 \right) \left( 5 y ^ { 2 } + 4 \right)$
C) $( 5 y - 4 ) ( 3 y + 2 )$
D) $\left( 3 y ^ { 2 } + 1 \right) \left( 5 y ^ { 2 } - 8 \right)$

Question

Quizplus · Accepted Answer

The ac-method involves multiplying the coefficient of the $y^4$ term (15) by the constant term (-8), giving -120. We look for two numbers that multiply to -120 and add to the coefficient of the $y^2$ term (2). Those numbers are -10 and 12. We then rewrite the middle term (2$y^2$) as -10$y^2$ + 12$y^2$ and factor by grouping, leading to the correct factorization $(3y^2 - 2)(5y^2 + 4)$.

Factor Using the Ac-Method 15y4+2y2−815 y ^ { 4 } + 2 y ^ { 2 } - 815y4+2y2−8

Factor Using the Ac-Method $15 y ^ { 4 } + 2 y ^ { 2 } - 8$